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A flask is filled with equal moles of $\mathrm{A~and ~B}$ . The half lives of $\mathrm{A ~and ~B ~are ~100 \mathrm{~s} ~and ~50 \mathrm{~s}}$ respectively and are independent of the initial concentration. The time required for the concentration of $\mathrm{A}$ to be four times that of $\mathrm{B}$ is ______ $\mathrm{s} .$

(Given : $\ln 2=0.693$ )
Option: 1
200

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

Given , half Life of A, $\mathrm{\left(t_{\frac{1}{2}}\right)_{A}=100\, \mathrm{sec}}$
half life of B , $\mathrm{\left(t_{\frac{1}{2}}\right)_{B}=50\, \mathrm{sec}}$

Since, initial concentration is independent of half life signifies that Reaction is of first order.

$\mathrm{\therefore\; \text{for first order Reaction,}\; t_{\frac{1}{2}}=\frac{\ln 2}{k} }$ .
$\mathrm{K_{A}=\frac{\ln 2 }{100},\; K_{B}=\frac{\ln 2}{50}}$

For I^storder $\mathrm{R\alpha ^{n}\rightarrow A= A_{0}\, e^{-kt}}$

$\mathrm{For \: A \rightarrow A_{t}=A_{0} \; e^{\left(-\frac{\ln 2}{100} t\right)}}$
$\mathrm{For \; B \rightarrow B_{t}=B_{0} \; e^{\left(-\frac{\ln 2}{50} t\right)}}$

Since initial concentration of $\mathrm{A}$ and $\mathrm{B}$ are same.
$\mathrm{\text { So } A_{0}=B_{0} \quad -{ (1) }}$
and condition given is -
$\mathrm{A_{t}= 4B_t \; \quad -(2)}$

Putting Both equation (1), (2) we get -

$\mathrm{e^{-\frac{\ln 2}{100} t}=4 e^{-\frac{\ln 2}{50} t}$